Kragujevac Journal of Mathematics
Volume 37(2) (2013), Pages 257–268.
THE RAINBOW DOMINATION NUMBER OF A DIGRAPH
J. AMJADI1 , A. BAHREMANDPOUR1 , S. M. SHEIKHOLESLAMI1 , AND L. VOLKMANN2
Abstract. Let D = (V, A) be a finite and simple digraph. A II-rainbow dominating function (2RDF) of a digraph D is a function f from the vertex set V to the
set of all subsets
of the set {1, 2} such that for any vertex v ∈ V with f (v) = ∅ the
S
condition u∈N − (v) f (u) = {1, 2} is fulfilled, where N − (v) is the set of in-neighbors
P
of v. The weight of a 2RDF f is the value ω(f ) = v∈V |f (v)|. The 2-rainbow
domination number of a digraph D, denoted by γr2 (D), is the minimum weight of a
2RDF of D. In this paper we initiate the study of rainbow domination in digraphs
and we present some sharp bounds for γr2 (D).
1. Introduction
Let D be a finite simple digraph with vertex set V (D) = V and arc set A(D) = A. A
digraph without directed cycles of length 2 is an oriented graph. The order n = n(D)
of a digraph D is the number of its vertices. We write d+
D (v) for the outdegree of
−
a vertex v and dD (v) for its indegree. The minimum and maximum indegree and
minimum and maximum outdegree of D are denoted by δ − = δ − (D), ∆− = ∆− (D),
δ + = δ + (D) and ∆+ = ∆+ (D), respectively. If uv is an arc of D, then we also write
u → v, and we say that v is an out-neighbor of u and u is an in-neighbor of v. For
a vertex v of a digraph D, we denote the set of in-neighbors and out-neighbors of v
by N − (v) = ND− (v) and N + (v) = ND+ (v), respectively. LetSN − [v] = N − (v) ∪ {v} and
N + [v] = N + (v) ∪ {v}. For S ⊆ V (D), we define N + [S] = v∈S N + [v]. If X ⊆ V (D),
then D[X] is the subdigraph induced by X. If X ⊆ V (D) and v ∈ V (D), then E(X, v)
is the set of arcs from X to v. Consult [2, 7] for the notation and terminology which
are not P
defined here. For a real-valued function f : VP
(D) −→ R the weight of f is
w(f ) = v∈V f (v), and for S ⊆ V , we define f (S) = v∈S f (v), so w(f ) = f (V ).
A vertex v dominates all vertices in N + [v]. A subset S of vertices of D is a dominating set if S dominates V (D). The domination number γ(D) is the minimum
Key words and phrases. Rainbow dominating function, Rainbow domination number, Digraph.
2010 Mathematics Subject Classification. Primary: 05C69. Secondary: 05C20.
Received: July 5, 2013.
257
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J. AMJADI, A. BAHREMANDPOUR, S. M. SHEIKHOLESLAMI, AND L. VOLKMANN
cardinality of a dominating set of D. The domination number of digraphs was introduced by Chartrand, Harary and Yue [1] as the out-domination number and has been
studied by several authors (see, for example [4, 8]). A Roman dominating function
(RDF) on a digraph D is a function f : V −→ {0, 1, 2} satisfying the condition that
every vertex v for which f (v) = 0 has P
an in-neighbor u for which f (u) = 2. The
weight of an RDF f is the value ω(f ) = v∈V f (v). The Roman domination number
of a digraph D, denoted by γR (D), equals the minimum weight of an RDF on D.
A γR (D)-function is a Roman dominating function of D with weight γR (D). The
Roman domination number in digraphs was introduced by Kamaraj and Jakkammal
[3] and has been studied in [5]. A Roman dominating function f : V −→ {0, 1, 2} can
be represented by the ordered partition (V0 , V1 , V2 ) (or (V0f , V1f , V2f ) to refer f ) of V ,
where Vi = {v ∈ V | f (v) = i}.
For a positive integer k, a k-rainbow dominating function (kRDF) of a digraph D is
a function f from the vertex set V (D) to the set of all subsets of theSset {1, 2, . . . , k}
such that for any vertex v ∈ V (D) with f (v) = ∅ the condition u∈N − (v) f (u) =
P
{1, 2, . . . , k} is fulfilled. The weight of a kRDF f is the value ω(f ) = v∈V |f (v)|.
The k-rainbow domination number of a digraph D, denoted by γrk (D), is the minimum weight of a kRDF of D. A γrk (D)-function is a k-rainbow dominating function of D with weight γrk (D). Note that γr1 (D) is the classical domination number γ(D). A 2-rainbow dominating function (briefly, rainbow dominating function)
f : V −→ P({1, 2}) can be represented by the ordered partition (V0 , V1 , V2 , V1,2 ) (or
f
(V0f , V1f , V2f , V1,2
) to refer f ) of V , where V0 = {v ∈ V | f (v) = ∅}, V1 = {v ∈ V |
f (v) = {1}}, V2 = {v ∈ V | f (v) = {2}} and V1,2 = {v ∈ V | f (v) = {1, 2}}. In
this representation, its weight is ω(f ) = |V1 | + |V2 | + 2|V1,2 |. Since V1 ∪ V2 ∪ V1,2 is
a dominating set when f is a 2RDF, and since assigning {1, 2} to the vertices of a
dominating set yields an 2RDF, we have
(1.1)
γ(D) ≤ γr2 (D) ≤ 2γ(D).
Our purpose in this paper is to establish some bounds for the rainbow domination
number of a digraph.
For S ⊆ V (D), the S-out-private neighbors of a vertex v of S are the vertices of
N + [v] \ N + [S − {v}]. The vertex v is its own out-private neighbor if v 6∈ N + [S − {v}].
The other out-private neighbors are external, i.e., belong to V − S. We make use of
the following result in this paper.
Proposition 1.1. [3] Let f = (V0 , V1 , V2 ) be any γR (D)-function of a digraph D.
Then
(a) If w ∈ V1 , then ND− (w) ∩ V2 = ∅.
(b) Let H = D[V0 ∪ V2 ]. Then each vertex v ∈ V2 with N − (v) ∩ V2 6= ∅, has at
least two out-private neighbors relative to V2 in the digraph H.
Proposition 1.2. Let k ≥ 1 be an integer. If D is a digraph of order n, then
min{k, n} ≤ γrk (D) ≤ n.
THE RAINBOW DOMINATION NUMBER OF A DIGRAPH
259
In particular, γrn (D) = n.
Proof. Let f be a γrk (D)-function. If there exists a vertex v such that f (v) = ∅,
then the definition yields to f (N − (v)) = {1, 2, . . . , k} and thus k ≤ γrk (D). If f (v)
is nonempty for all vertices v ∈ V (D), then n ≤ γrk (D), and the first inequality is
proved.
Next consider the function g, defined by g(v) = {1} for each v ∈ V (D). Then g is
a k-rainbow dominating function of weight n, and so γrk (D) ≤ n.
Proposition 1.3. Let k ≥ 1 be an integer. If D is a digraph of order n, then
γrk (D) ≤ n − ∆+ (D) + k − 1.
Proof. Let v be a vertex of maximum outdegree ∆+ (D). Define f : V (D) →
P({1, 2, . . . , k}) by f (v) = {1, 2, . . . , k}, f (x) = ∅ if x ∈ N + (v) and f (x) = {1}
otherwise. It is easy to see that f is a k-rainbow dominating function of D and thus
γrk (D) ≤ n − ∆+ (D) + k − 1.
Let k ≥ 1 be an integer, and let D be a digraph of order n ≥ k and maximum
outdegree ∆+ (D) = n − 1. Since n ≥ k, Proposition 1.2 leads to γrk (D) ≥ k. Hence
it follows from Proposition 1.3 that
k ≤ γrk (D) ≤ n − ∆+ (D) + k − 1 = k
and therefore γrk (D) = k. This example shows that Proposition 1.3 is sharp.
2. Bounds on the rainbow domination number of digraphs
Theorem 2.1. For a digraph D, 32 γR (D) ≤ γr2 (D) ≤ γR (D).
Proof. The upper bound is immediate by definition. To prove the lower bound, let
f be a γr2 (D)-function and let Xi = {v ∈ V (D) | i ∈ f (v)} for i = 1, 2. We
may assume that |X1 | ≤ |X2 |. Then |X1 | ≤ (|X1 | + |X2 |)/2 = γr2 (D)/2. Define
g : V (D) → {0, 1, 2} by g(u) = 0 if f (u) = ∅, g(u) = 1 when f (u) = {2} and
g(u) = 2 if 1 ∈ f (u). Obviously, g is a Roman dominating function on D with
ω(g) ≤ 2|X1 | + |X2 | ≤ 23 γr2 (D) and the result follows.
Lemma 2.1. Let k ≥ 0 be an integer and let D be a digraph of order n.
(i) If γR (D) = γr2 (D) + k, then there exists a subset U of V (D) such that
|N + [U ]| = n − γR (D) + 2|U |. In particular, there exists a subset U of V (D)
such that |N + [U ]| = n − γr2 (D) + 2|U | − k.
(ii) For each U ⊆ V (D), γR (D) ≤ n − (|N + [U ]| − 2|U |). In particular, if there
exists a subset U of V (D) such that |N + [U ]| = n − γr2 (D) + 2|U | − k, then
γR (D) ≤ γr2 (D) + k.
(iii) If U ∗ is a subset of V (D) such that |N + [U ∗ ]|−2|U ∗ | is maximum, then γR (D) =
n − (|N + [U ∗ ]| − 2|U ∗ |).
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J. AMJADI, A. BAHREMANDPOUR, S. M. SHEIKHOLESLAMI, AND L. VOLKMANN
Proof. (i) Let f = (V0 , V1 , V2 ) be a γR (D)-function. By Proposition 1.1 (a), N + [V2 ] =
V0 ∪ V2 . Since γR (D) = |V1 | + 2|V2 |, we have
|N + [V2 ]| = |V0 | + |V2 | = (n − |V1 | − |V2 |) + |V2 | = n − (γR (D) − 2|V2 |).
Hence V2 is the desired subset of V (D).
(ii) Let U be a subset of V (D). Clearly, f = (N + [U ] − U, V (D) − N + [U ], U ) is an
RDF of D of weight n − (|N + [U ]| − 2|U |) and hence γR (D) ≤ n − (|N + [U ]| − 2|U |).
(iii) It follows from (i) and the choice of U ∗ that
|N + [U ∗ ]| − 2|U ∗ | ≥ n − γr2 (D) − k = n − γR (D),
when γR (D) = γr2 (D) + k.
By (ii), we obtain
γR (D) ≤ n − (|N + [U ∗ ]| − 2|U ∗ |) ≤ n − (n − γR (D)) = γR (D)
and the proof is complete.
Theorem 2.2. Let k ≥ 0 be an integer and let D be a digraph of order n. Then
γR (D) = γr2 (D) + k if and only if
(i) there exists no subset U of V (D) such that
n − γr2 (D) + 2|U | − k + 1 ≤ |N + [U ]| ≤ n − γr2 (D) + 2|U | and
(ii) there exists a subset U of V (D) such that |N + [U ]| = n − γr2 (D) + 2|U | − k.
Proof. The proof is by induction on k. Let k = 0. If γR (D) = γr2 (D), then (i)
is trivial and (ii) follows from Lemma 2.1 (i). Now assume that (i) and (ii) hold.
It follows from Lemma 2.1 (ii) that γR (D) ≤ γr2 (D). By Theorem 2.1, we have
γR (D) = γr2 (D), as desired. Therefore we may assume that k ≥ 1 and the theorem
is true for each m ≤ k − 1.
Let γR (D) = γr2 (D) + k. By Lemma 2.1 (i), it suffices to show that (i) holds.
Assume to the contrary that there exists an integer 0 ≤ i ≤ k − 1 and a subset
U ⊆ V (D) such that n − γr2 (D) + 2|U | − i ≤ |N + [U ]|. By choosing (U, i) so that i is
as small as possible and by the inductive hypothesis we have γR (D) ≤ γr2 (D) + i <
γr2 (D) + k which is a contradiction. This proves the “only” part of the theorem.
Conversely, assume that (i) and (ii) hold. By Lemma 2.1 (ii) we need to show that
γR (D) ≥ γr2 (D) + k. Suppose i = γR (D) − γr2 (D). If 0 ≤ i ≤ k − 1, then by the
inductive hypothesis there exists a set U ⊆ V (D) such that n − γr2 (D) + 2|U | − i =
|N + [U ]| which contradicts (i). Hence γR (D) = γr2 (D) + k and the proof is complete.
Theorem 2.3. Let D be a digraph of order n. Then
2n
γr2 (D) ≥
.
2 + ∆+ (D)
THE RAINBOW DOMINATION NUMBER OF A DIGRAPH
261
Proof. Let f = (V0 , V1 , V2 , V1,2 ) be a γr2 (D)-function. Then γr2 (D) = |V1 | + |V2 | +
2|V1,2 | and n = |V0 | + |V1 | + |V2 | + |V1,2 |.
Since each vertex of V0 has at least one in-neighbor in V1,2 or at least one in-neighbor
in V2 and at least one in-neighbor in V1 , we deduce that |V0 | ≤ ∆+ (|V2 | + |V1,2 |) and
|V0 | ≤ ∆+ (|V1 | + |V1,2 |). Hence, we conclude that
(∆+ + 2)γr2 (D) = (∆+ + 2)(|V1 | + |V2 | + 2|V1,2 |)
= (∆+ + 2)(|V1 | + |V2 |) + 4|V1,2 | + 2∆+ |V1,2 |
≥ 2|V1 | + 2|V2 | + 2|V0 | + 4|V1,2 |
= 2n + 2|V1,2 |
≥ 2n,
as desired.
The proof of Theorem 2.3 shows that
2n + 2
γr2 (D) ≥
2 + ∆+ (D)
f
if there exists a γ2r (D)-function f such that V1,2
6= ∅. Let G be a graph and γr2 (G)
its 2-rainbow domination number. If G is of order n and maximum degre ∆, then
Theorem 2.3 implies immediately the known bound γr2 (G) ≥ d2n/(∆ + 2)e, given
by Sheikhoelslami and Volkmann [6]. Next we characterize the digraphs D with
γr2 (D) = 2.
Proposition 2.1. Let D be a digraph of order n ≥ 2. Then γr2 (D) = 2 if and only
if n = 2 or n ≥ 3 and ∆+ (D) = n − 1 or there exist two different vertices u and v
such that V (D) − {u, v} ⊆ N + (u) and V (D) − {u, v} ⊆ N + (v).
Proof. If n = 2 or n ≥ 3 and ∆+ (D) = n − 1 or there exist two different vertices u
and v such that V (D) − {u, v} ⊆ N + (u) and V (D) − {u, v} ⊆ N + (v), then it is easy
to see that γr2 (D) = 2.
Conversely, assume that γr2 (D) = 2. Let f = (V0 , V1 , V2 , V1,2 ) be a γr2 (D)-function.
Clearly, 2 = γr2 (D) = |V1 | + |V2 | + 2|V1,2 | and thus |V1,2 | ≤ 1. If |V1,2 | = 1, then
|V1 | = |V2 | = 0 and hence ∆+ (D) = n − 1. If |V1,2 | = 0, then |V1 |, |V2 | ≤ 2. If |V1 | = 0
or |V2 | = 0, then we deduce that n = 2. In the remaining case that |V1 | = |V2 | = 1,
we assume that V1 = {u} and V2 = {v}. The definition of the 2-rainbow dominating
function implies that each vertex of V (D) − {u, v} has u and v as an in-neighbor.
Consequently, V (D) − {u, v} ⊆ N + (u) and V (D) − {u, v} ⊆ N + (v).
Proposition 2.2. Let D be a digraph of order n. Then γr2 (D) < n if and only if
∆+ (D) ≥ 2 or ∆− (D) ≥ 2.
Proof. Let f = (V0 , V1 , V2 , V1,2 ) be a γr2 (D)-function of D. The hypothesis |V0 | +
|V1 | + |V2 | + |V1,2 | = n > γr2 (D) = |V1 | + |V2 | + 2|V1,2 | implies |V0 | ≥ |V1,2 | + 1. If some
vertex w ∈ V0 has no in-neighbor in V1,2 , then |N − (w) ∩ V1 | ≥ 1 and |N − (w) ∩ V2 | ≥ 1
262
J. AMJADI, A. BAHREMANDPOUR, S. M. SHEIKHOLESLAMI, AND L. VOLKMANN
which implies that ∆− (D) ≥ d− (w) ≥ 2. So we may assume each vertex w ∈ V0 has
at least one in-neighbor in V1,2 . Then we have
X
d+
D (u) ≥ |V0 | ≥ |V1,2 | + 1.
u∈V1,2
If we suppose on the contrary that ∆+ (D) ≤ 1, then we arrive at the contradiction
X
|V1,2 | ≥
d+
D (u) ≥ |V1,2 | + 1.
u∈V1,2
+
If ∆ (D) ≥ 2, then Proposition 1.3 implies that γr2 (D) ≤ n − ∆+ (D) + 1 < n. If
∆− (D) ≥ 2, then assume that u is a vertex with in-degree ∆− (D) and let v, w ∈
N − (u) be two distinct vertices. Then ({u}, V (D) − {u, v}, {v}, ∅) is a rainbow dominating function of D implying that γr2 (D) < n, and the proof is complete.
Corollary 2.1. If D is a directed path or directed cycle of order n, then γr2 (D) = n.
Next we characterize the digraphs which attain the lower bound in (1.1).
Proposition 2.3. Let D be a digraph on n vertices. Then γ(D) = γr2 (D) if and
only if D has a γ(D)-set S that partitions into two nonempty subsets S1 and S2 such
that N + (S1 ) = V (D) − (S1 ∪ S2 ) and N + (S2 ) = V (D) − (S1 ∪ S2 ).
Proof. Assume that γ(D) = γr2 (D) and let f = (V0 , V1 , V2 , V1,2 ) be a γr2 -function
of D. If γ(D) = γr2 (D) = n, then clearly D is empty and the result is immediate.
Let γ(D) = γr2 (D) < n. Then the assumption implies that we have equality in
γ(D) ≤ |V1 | + |V2 | + |V1,2 | ≤ |V1 | + |V2 | + 2|V1,2 | = γr2 (D). This implies that |V1,2 | = 0
and hence we deduce that each vertex in V0 has at least one in-neighbor in V1 and
one in-neighbor in V2 . Therefore, V (D) − (V1 ∪ V2 ) ⊆ N + (V1 ) and V (D) − (V1 ∪ V2 ) ⊆
N + (V2 ). If there is an arc (a, b) in D[V1 ∪ V2 ], then obviously (V1 ∪ V2 ) − {b} is a
dominating set of D which is a contradiction. Hence V1 ∪ V2 is dominating set of D
with V (D) − (V1 ∪ V2 ) = N + (V1 ) and V (D) − (V1 ∪ V2 ) = N + (V2 ).
Conversely, assume that D has a minimum dominating set S that partitions into two
nonempty subsets S1 and S2 such that N + (S1 ) = V (D) − (S1 ∪ S2 ) and N + (S2 ) =
V (D) − (S1 ∪ S2 ). It is straightforward to verify that the function (V (D) − (S1 ∪
S2 ), S1 , S2 , ∅) is a rainbow dominating function of D of weight γ(D) and hence γ(D) =
γr2 (D). This completes the proof.
3. Cartesian product of directed cycles
Let D1 = (V1 , A1 ) and D2 = (V2 , A2 ) be two digraphs which have disjoint vertex
sets V1 and V2 and disjoint arc sets A1 and A2 , respectively. The Cartesian product
D1 D2 is the digraph with vertex set V1 × V2 and for any two vertices (x1 , x2 ) and
(y1 , y2 ) of D1 D2 , (x1 , x2 )(y1 , y2 ) ∈ A(D1 D2 ) if one of the following holds:
(i) x1 = y1 and x2 y2 ∈ A(D2 );
(ii) x1 y1 ∈ A(D1 ) and x2 = y2 .
THE RAINBOW DOMINATION NUMBER OF A DIGRAPH
263
We denote the vertices of a directed cycle Cn by the integers {1, 2, . . . , n} considered
modulo n. Note that N + ((i, j)) = {(i, j + 1), (i + 1, j)} for any vertex (i, j) ∈
V (Cm Cn ), the first and second digit are considered modulo m and n, respectively.
k
the subdigraph of Cm Cn induced by
For any k ∈ {1, 2, . . . , n}, we will denote by Cm
k
the vertices {(j, k)|j ∈ {1, 2, . . . , m}}.
PNote that Cm is isomorphic to Cm . Let f be a
γr2 (Cm Cn )-function and set ak = x∈V (Cmk ) |f (x)| for any k ∈ {1, 2, . . . , n}. Then
P
γr2 (Cm Cn ) = nk=1 ak . It is easy to see that Cm Cn ∼
= Cn Cm for any directed
cycles of length m, n ≥ 2, and so γr2 (Cm Cn ) = γr2 (Cn Cm ).
n
if n is even
Theorem 3.1. For n ≥ 2, γr2 (C2 Cn ) =
n + 1 if n is odd.
Proof. First let n be even. Define f : V (C2 Cn ) → P({1, 2}) by f ((1, 2i − 1)) = {1}
for 1 ≤ i ≤ n2 , f ((2, 2i)) = {2} for 1 ≤ i ≤ n2 and f (x) = ∅ otherwise. Obviously, f is
a 2RDF of C2 Cn of weight n and hence γr2 (C2 Cn ) ≤ n. On the other hand, since
∆+ (C2 Cn ) = 2, it follows from Theorem 2.3 that γr2 (C2 Cn ) = n.
Now let n be odd. We claim that γr2 (C2 Cn ) ≥ n+1. Assume to
Pthe contrary that
γr2 (C2 Cn ) ≤ n. Let f be a γr2 (C2 Cn )-function and set ak = x∈V (C k ) |f (x)| for
2
any k ∈ {1, 2, . . . , n}. If ak = 0 for some k, say k = 3, then f ((1, 3)) = f ((2, 3)) = ∅
and to dominate the vertices (1, 3) and (2, 3) we must have f ((1, 2)) = {1, 2} and
f ((2, 2)) = {1, 2}, respectively. Then the function g : V (C2 Cn ) → P({1, 2}) by
g((1, 2)) = {1}, g((2, 2)) = ∅, g((2, 1)) = g((2, 3)) = {2} and g(x) = f (x) for x ∈
V (C2 Cn ) − {(1, 2), (2, 2), (2, 1), (2, 3)} is a 2RDF of C2 Cn of weight less than ω(f )
which is a contradiction. Thus ak ≥ 1 for each k. By assumption ak = 1 for each k.
We may assume, without loss of generality, that f ((1, 2)) = {1}. To dominate (2, 2),
we must have f ((2, 1)) = {2}. Since a3 = 1 and f ((2, 2)) = ∅, to dominate ((1, 3)),
we must have f ((2, 3)) = {2}. Repeating this process we obtain f ((1, 2i)) = {1} for
1 ≤ i ≤ n−1
and f ((2, 2i − 1)) = {2} for 1 ≤ i ≤ n+1
and f (x) = ∅ otherwise. But
2
2
then the vertex (1, 1) is not dominated, a contradiction. Thus γr2 (C2 Cn ) ≥ n + 1.
Define g : V (C2 Cn ) → P({1, 2}) by g((1, 1)) = {1}, g((1, 2i)) = {1} for 1 ≤ i ≤ n−1
2
and g((2, 2i − 1)) = {2} for 1 ≤ i ≤ n+1
and g(x) = ∅ otherwise. Clearly, g is a 2RDF
2
of C2 Cn of weight n + 1 and hence γr2 (C2 Cn ) = n + 1.
Theorem 3.2. For n ≥ 2, γr2 (C3 Cn ) = 2n.
Proof. First we prove γr2 (C3 Cn ) ≤ 2n. Define g : V (C3 Cn ) → P({1, 2}) as follows:
If n ≡ 0 (mod 3), then g((1, 3i+1)) = g((2, 3i+2)) = g((3, 3i+3)) = {1}, g((1, 3i+
3)) = g((2, 3i + 1)) = g((3, 3i + 2)) = {2} for 0 ≤ i ≤ n3 − 1 and g(x) = ∅ otherwise,
if n ≡ 1 (mod 3), then g((3, n))) = {1}, g((2, n)) = {2}, g((1, 3i + 1)) = g((2, 3i +
2)) = g((3, 3i + 3)) = {1}, g((1, 3i + 3)) = g((2, 3i + 1)) = g((3, 3i + 2)) = {2} for
0 ≤ i ≤ n−1
− 1 and g(x) = ∅ otherwise,
3
if n ≡ 2 (mod 3), then g((1, n)) = g((1, n − 1)) = g((3, n)) = {1}, g((2, n − 1)) =
{2}, g((1, 3i+1)) = g((2, 3i+2)) = g((3, 3i+3)) = {1}, g((1, 3i+3)) = g((2, 3i+1)) =
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J. AMJADI, A. BAHREMANDPOUR, S. M. SHEIKHOLESLAMI, AND L. VOLKMANN
g((3, 3i + 2)) = {2} for 0 ≤ i ≤ n−2
− 1 and g(x) = ∅ otherwise. It is easy to see that
3
in each case, g is a 2RDF of C3 Cn of weight 2n and hence γr2 (C3 Cn ) ≤ 2n.
Now we P
show that γr2 (C3 Cn ) ≥ 2n. First we prove that for P
any γr2 (C3 Cn )function f , x∈V (Cmk ) |f (x)| ≥ 1 for each k. Let to the contrary that x∈V (Cmk ) |f (x)| =
0 for some k, say k = n. Then we must have f ((1, n − 1)) = f ((2, n − 1)) =
f ((3, n−1)) = {1, 2}. Then the function f1 defined by f1 ((1, n−1)) = f1 ((2, n−1)) =
f1 ((3, n − 1)) = {1}, f1 ((1, n)) = f1 ((2, n)) = {2} and f1 (x) = f (x) otherwise, is a
2RDF of G of weight less than ω(f ), a contradiction.
PLet g be a γr2 (C3 Cn )-function such that the size of the set S = {k | ak =
k ) |g(x)| = 1 and 1 ≤ i ≤ n} is minimum. We claim that |S| = 0 implying that
x∈V (Cm
γr2 (C3 Cn ) = ω(g) ≥ 2n. Assume to the contrary that |S| ≥ 1. Suppose, without
loss of generality, that an = 1 and that g((1, n)) = {1} and g((2, n)) = g((3, n)) = ∅.
To dominate (2, n) and (3, n) we must have 2 ∈ g((2, n−1)) and g((3, n−1)) = {1, 2},
respectively. If n = 3, then clearly a1 ≥ 2 and so γr2 (C3 Cn ) ≥ 2n. Let n ≥ 4.
Consider two cases.
Case 1. an−2 = 1.
As above we have an−3 ≥ 3. Thus an−3 + an−2 + an−1 + an ≥ 8. If n = 4, we are
done. Suppose n ≥ 5. Since an−4 ≥ 1, 1 ∈ ∪3i=1 g((i, n − 4)) or 2 ∈ ∪3i=1 g((i, n − 4)).
Let 1 ∈ ∪3i=1 g((i, n − 4)) (the case 2 ∈ ∪3i=1 g((i, n − 4)) is similar).
If 1 ∈ g((3, n − 4)), then define h : V (C3 Cn ) → P({1, 2}) by h((1, n − 3)) =
h((1, n − 1)) = h((2, n − 3)) = h((3, n − 1)) = {2}, h((1, n)) = h((2, n)) = h((2, n −
2)) = h((3, n − 2)) = {1}, h((1, n − 2)) = h((2, n − 1)) = h((3, n − 3)) = h((3, n)) = ∅
and h(x) = g(x) otherwise.
If 1 ∈ g((2, n − 4)), then define h : V (C3 Cn ) → P({1, 2}) by h((1, n − 3)) =
h((2, n − 1)) = h((3, n − 1)) = h((3, n − 3)) = {2}, h((1, n)) = h((1, n − 2)) = h((2, n −
2)) = h((3, n)) = {1}, h((2, n − 2)) = h((1, n − 1)) = h((2, n − 3)) = h((2, n)) = ∅ and
h(x) = g(x) otherwise.
If 1 ∈ g((1, n − 4)), then define h : V (C3 Cn ) → P({1, 2}) by h((2, n − 3)) =
h((2, n − 1)) = h((3, n − 3)) = h((3, n − 1)) = {2}, h((1, n)) = h((1, n − 2)) = h((3, n −
2)) = h((3, n)) = {1}, h((2, n − 2)) = h((1, n − 1)) = h((1, n − 3)) = h((2, n)) = ∅ and
h(x) = g(x) otherwise.
P
Clearly, h is a 2RDF of C3 Cn for which |{k | x∈V (Cmk ) |h(x)| = 1 and 1 ≤ i ≤
P
n}| < |{k | x∈V (Cmk ) |g(x)| = 1 and 1 ≤ i ≤ n}| which contradicts the choice of g.
Case 2. an−2 ≥ 2.
Then an−2 +an−1 +an ≥ 6. Since an−3 ≥ 1, 1 ∈ ∪3i=1 g((i, n−3)) or 2 ∈ ∪3i=1 g((i, n−3)).
Assume 1 ∈ ∪3i=1 g((i, n − 3)) (the case 2 ∈ ∪3i=1 g((i, n − 3)) is similar).
If 1 ∈ g((3, n − 3)), then define h : V (C3 Cn ) → P({1, 2}) by h((1, n − 2)) =
h((2, n−2)) = h((2, n−1)) = {2}, h((1, n)) = h((3, n)) = h((3, n−1)) = {1}, h((1, n−
1)) = h((2, n)) = h((3, n − 2)) = ∅ and h(x) = g(x) otherwise.
THE RAINBOW DOMINATION NUMBER OF A DIGRAPH
265
If 1 ∈ g((2, n − 3)), then define h : V (C3 Cn ) → P({1, 2}) by h((1, n − 2)) =
h((2, n−1)) = h((3, n−2)) = {2}, h((1, n)) = h((3, n)) = h((3, n−1)) = {1}, h((1, n−
1)) = h((2, n − 2)) = h((2, n)) = ∅ and h(x) = g(x) otherwise.
If 1 ∈ g((1, n − 3)), then define h : V (C3 Cn ) → P({1, 2}) by h((2, n − 2)) =
h((3, n−2)) = h((3, n−1)) = {2}, h((1, n)) = h((1, n−1)) = h((2, n)) = {1}, h((3, n)) =
h((1, n − 2)) = h((2, n − 1)) = ∅ and h(x) = g(x) otherwise.
P
Clearly, h is a 2RDF of C3 Cn for which |{k | x∈V (Cmk ) |h(x)| = 1 and 1 ≤ i ≤
P
n}| < |{k | x∈V (Cmk ) |g(x)| = 1 and 1 ≤ i ≤ n}| which is a contradiction again.
P
Therefore, |{k | ak =
k ) |g(x)| = 1 and 1 ≤ i ≤ n}| = 0 and hence
x∈V (Cm
γr2 (C3 Cn ) = ω(g) ≥ 2n. Thus γr2 (C3 Cn ) = 2n and the proof is complete.
Proposition 3.1. If m = 2r and n = 2s for some positive integers r, s, then
γr2 (Cm Cn ) = mn
.
2
Proof. Define f : V (Cm Cn ) → P({1, 2}) by f ((2i − 1, 2j − 1)) = {1} for each
1 ≤ i ≤ r and 1 ≤ j ≤ s, f ((2i, 2j)) = {2} for each 1 ≤ i ≤ r and 1 ≤ j ≤ s and
f (x) = ∅ otherwise. It is easy to see that f is a 2RDF of Cm Cn with weight mn
and
2
so γr2 (Cm Cn ) ≤ mn
.
Now
the
results
follows
from
Theorem
2.3.
2
4. Strong product of directed cycles
Let D1 = (V1 , A1 ) and D2 = (V2 , A2 ) be two digraphs which have disjoint vertex
sets V1 and V2 and disjoint arc sets A1 and A2 , respectively. The strong product
D1 ⊗ D2 is the digraph with vertex set V1 × V2 and for any two vertices (x1 , x2 ) and
(y1 , y2 ) of D1 ⊗ D2 , (x1 , x2 )(y1 , y2 ) ∈ A(D1 ⊗ D2 ) if one of the following holds:
(i) x1 y1 ∈ A(D1 ) and x2 y2 ∈ A(D2 );
(ii) x1 = y1 and x2 y2 ∈ A(D2 );
(iii) x1 y1 ∈ A(D1 ) and x2 = y2 .
We denote the vertices of a directed cycle Cn by the integers {1, 2, . . . , n} considered
modulo n. There is an arc xy from x to y in Cn if and only if y = x + 1 (mod n) and
N + ((i, j)) = {(i, j + 1), (i + 1, j), (i + 1, j + 1)} for any vertex (i, j) ∈ V (Cm ⊗ Cn ),
the first and second digit are considered modulo m and n, respectively. We use the
notation defined in Section 3.
Lemma 4.1. For positive integers m, n ≥ 2, γr2 (Cm ⊗ Cn ) ≥ d mn
e.
2
k
k−1
k
Proof. Observe that the vertices of Cm
are dominated by vertices of Cm
or Cm
,
1
1
n
k = 1, 2, . .P
. , n. Especially, the vertices of Cm are dominated by Cm and Cm . We
show that nk=1 ak ≥ d mn
e. In order to this, we show that ak + ak+1 ≥ m for each
2
k = 1, 2, . . . , n, where an+1 = a1 . First let ak+1 = 0. Then to rainbowly dominate
(i, k + P
1) for each 1 ≤ i ≤ m, we must have |f ((i − 1, k))| + |f ((i, k))| ≥ 2. Then
2ak = m
i=1 (|f ((i − 1, k))| + |f ((i, k))|) ≥ 2m and hence ak + ak+1 ≥ m. If ak+1 = t,
266
J. AMJADI, A. BAHREMANDPOUR, S. M. SHEIKHOLESLAMI, AND L. VOLKMANN
then it is not hard to see that ak ≥ m − t and so ak + ak+1 ≥ m. Therefore,
n
n
X
X
2γr2 (Cm ⊗ Cn ) = 2
ak =
(ak + ak+1 ) ≥ nm
k=1
k=1
e.
where an+1 = a1 . This implies that γr2 (Cm ⊗ Cn ) ≥ d mn
2
Proposition 4.1. If m = 2r and n = 2s for some positive integers r, s, then γr2 (Cm ⊗
Cn ) = d mn
e.
2
Proof. Define f : V (D) → P({1, 2}) by f ((2i − 1, 2j − 1)) = {1, 2} for each 1 ≤ i ≤ r
and 1 ≤ j ≤ s, and f (x) = ∅ otherwise. It is easy to see that f is a 2RDF of Cm ⊗ Cn
with weight mn
and so γr2 (Cm ⊗ Cn ) ≤ mn
. Now the results follows from Lemma
2
2
4.1.
Proposition 4.2. If m = 4r and n = 2s + 1 for some positive integers r, s, then
γr2 (Cm ⊗ Cn ) = d mn
e.
2
Proof. Let
W1 = {(4i + 1, 1) | 0 ≤ i ≤ r − 1},
W2 = {(4i + 3, 1) | 0 ≤ i ≤ r − 1},
Z11 = {(4i + 2, 2j) | 0 ≤ i ≤ r − 1 and 1 ≤ j ≤ s},
Z12 = {(4i + 4, 2j) | 0 ≤ i ≤ r − 1 and 1 ≤ j ≤ s},
Z21 = {(4i + 4, 2j + 1) | 0 ≤ i ≤ r − 1 and 1 ≤ j ≤ s},
Z22 = {(4i + 2, 2j + 1) | 0 ≤ i ≤ r − 1 and 1 ≤ j ≤ s}.
Define f : V (Cm ⊗ Cn ) → P({1, 2}) by f (x) = {1} for x ∈ W1 ∪ Z11 ∪ Z21 , f (x) = {2}
for x ∈ W2 ∪ Z12 ∪ Z22 , and f (x) = ∅ otherwise. It is easy to see that f is a 2RDF of
Cm ⊗ Cn with weight mn
and so γr2 (Cm ⊗ Cn ) ≤ mn
. It follows from Lemma 4.1 that
2
2
mn
γr2 (Cm ⊗ Cn ) = d 2 e.
Proposition 4.3. If m = 4r + 2 and n = 2s + 1 for some positive integers r, s, then
d mn
e ≤ γr2 (Cm ⊗ Cn ) ≤ d mn
e + 1.
2
2
Proof. Let
C1 = {(4r + 2, 2s + 1)} ∪ {(4i + 4, 2s + 1) | 0 ≤ i ≤ r − 1},
C2 = {(4r + 1, 2s + 1)} ∪ {(4i + 2, 2s + 1) | 0 ≤ i ≤ r − 1},
C1,2 = {(4r + 1, 2j + 1) | 0 ≤ j ≤ s − 1},
Z11 = {(4i + 1, 2j + 1) | 0 ≤ i ≤ r − 1 and 0 ≤ j ≤ s − 1},
Z12 = {(4i + 1, 2j) | 0 ≤ i ≤ r − 1 and 1 ≤ j ≤ s},
Z21 = {(4i + 3, 2j) | 0 ≤ i ≤ r − 1 and 1 ≤ j ≤ s},
Z22 = {(4i + 3, 2j + 1) | 0 ≤ i ≤ r − 1 and 0 ≤ j ≤ s − 1}.
THE RAINBOW DOMINATION NUMBER OF A DIGRAPH
267
Define f : V (Cm ⊗ Cn ) → P({1, 2}) by f (x) = {1, 2} for x ∈ C1,2 , f (x) = {1} for
x ∈ C1 ∪ Z11 ∪ Z21 , f (x) = {2} for x ∈ C2 ∪ Z12 ∪ Z22 , and f (x) = ∅ otherwise. It is easy
to see that f is a 2RDF of Cm ⊗ Cn with weight mn
+ 1 and so γr2 (Cm ⊗ Cn ) ≤ mn
+ 1.
2
2
mn
mn
It follows from Lemma 4.1 that d 2 e ≤ γr2 (Cm ⊗ Cn ) ≤ d 2 e + 1.
Lemma 4.2. If m = 4r + 3 and n = 4s + 3 for some non-negative integers r, s, then
e.
γr2 (Cm ⊗ Cn ) = d mn
2
Proof. Let
R11 = {(4i + 1, 4j + 1) | 0 ≤ i ≤ r and 0 ≤ j ≤ s},
R12 = {(4i + 1, 4j + 3) | 0 ≤ i ≤ r and 0 ≤ j ≤ s},
R21 = {(4i + 2, 4j + 2) | 0 ≤ i ≤ r and 0 ≤ j ≤ s},
R22 = {(4i + 2, 4j + 4) | 0 ≤ i ≤ r and 0 ≤ j ≤ s − 1},
R31 = {(4i + 3, 4j + 3) | 0 ≤ i ≤ r and 0 ≤ j ≤ s},
R32 = {(4i + 3, 4j + 1) | 0 ≤ i ≤ r and 0 ≤ j ≤ s},
R41 = {(4i + 4, 4j + 4) | 0 ≤ i ≤ r − 1 and 0 ≤ j ≤ s − 1},
R42 = {(4i + 4, 4j + 2) | 0 ≤ i ≤ r − 1 and 0 ≤ j ≤ s}.
Define f : V (Cm ⊗ Cn ) → P({1, 2}) by f (x) = {1} for x ∈ R11 ∪ R21 ∪ R31 ∪ R41 ,
f (x) = {2} for x ∈ R12 ∪ R22 ∪ R32 ∪ R42 , and f (x) = ∅ otherwise. It is easy to see that
f is a 2RDF of Cm ⊗ Cn with weight b mn
c + 1 and so γr2 (Cm ⊗ Cn ) ≤ d mn
e. By
2
2
mn
Lemma 4.1 we obtain γr2 (Cm ⊗ Cn ) = d 2 e.
Acknowledgment: This work has been supported by the Research Office of Azarbaijan Shahid Madani University.
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1
Department of Mathematics,
Azarbaijan Shahid Madani University,
Tabriz, I.R. Iran
E-mail address: [email protected]
E-mail address: [email protected]
2
Lehrstuhl II für Mathematik,
RWTH Aachen University,
52056 Aachen, Germany
E-mail address: [email protected]